How to customise Polar Plots

조회 수: 16 (최근 30일)
Sam Hurrell
Sam Hurrell 2023년 4월 2일
댓글: Star Strider 2023년 4월 6일
I have data typically graphed as 'value (B) against angle (A)' as a plot (for angles -90 to 90) that I wish to graph as a polar plot. What command(s) can I write that'll graph this data with polar axes for angle as: -90 (on the left) to 90 (on the right) via 0 (below in the middle)?
  댓글 수: 2
Star Strider
Star Strider 2023년 4월 2일
Some detail is missing.
How do you want the tick labels to appear on the left side?
What do you want the top value (currently 180°) to be?
th = linspace(0, 2*pi, 360);
r = sin(2*th).^2;
figure
polarplot(th,r)
Ax = gca;
Ax.ThetaZeroLocation = 'bottom';
TTL = Ax.ThetaTickLabel
TTL = 12×1 cell array
{'0°' } {'30°' } {'60°' } {'90°' } {'120°'} {'150°'} {'180°'} {'210°'} {'240°'} {'270°'} {'300°'} {'330°'}
.
Sam Hurrell
Sam Hurrell 2023년 4월 4일
Where you have 270 to 0 to 90, I want -90 to 0 to 90. I also want the top half of the graph gone leaving a semicircular polar plot.

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채택된 답변

Star Strider
Star Strider 2023년 4월 4일
Try something like this —
th = linspace(0, 2*pi, 360);
r = exp(-0.1*th).*sin(2*th).^2;
figure
polarplot(th,r)
title('Original')
figure
polarplot(th+pi/2,r) % It May Be Necessary To Shift The Plot Phase (+90° Or +π/2) For This To Work Correctly
Ax = gca;
Ax.ThetaZeroLocation = 'bottom';
Ax.ThetaLim = [270 270+180];
Ax.ThetaTickLabel = compose('%d°',-90:30:90);
title('Shifted')
I needed to make my original function radially asymmetric to test this to be certain it would work correctly. It may be necessary for you to experiment with shifting the angles as well.
.
  댓글 수: 3
Star Strider
Star Strider 2023년 4월 5일
One small thing though, how do I set custom gridlines (every 15^o rather than every 30^o)?
Try this —
th = linspace(0, 2*pi, 360);
r = exp(-0.1*th).*sin(2*th).^2;
figure
polarplot(th,r)
title('Original')
figure
polarplot(th+pi/2,r) % It May Be Necessary To Shift The Plot Phase (+90° Or +π/2) For This To Work Correctly
Ax = gca;
Ax.ThetaZeroLocation = 'bottom';
Ax.ThetaLim = [270 270+180];
Ax.ThetaTick = Ax.ThetaLim(1) : 15 : Ax.ThetaLim(2);
Ax.ThetaTickLabel = compose('%d°',-90:15:90);
title('Shifted With Ticks Every 15°')
.
Star Strider
Star Strider 2023년 4월 6일
If my Answer helped you solve your problem, please Accept it!
.

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추가 답변 (1개)

the cyclist
the cyclist 2023년 4월 2일
편집: the cyclist 2023년 4월 2일
% Create plot
figure
theta = linspace(0,2*pi,25);
rho = 2*theta;
polarplot(theta,rho);
% Define a variable with the tick labels
thetaTickLabels = {'0°';'30°';'60°';'90°';'120°';'150°';'180°';'-150°';'-120°';'-90°';'-60°';'-30°'};
% Set theta how you want
set(gca,"ThetaZeroLocation","bottom", ...
"ThetaDir","counterclockwise", ...
"ThetaTickLabels",thetaTickLabels)
See the documentation for polarplot for details.
Note that while the axes are labeled in degrees, the default input is in radians. (That same page shows how to convert, if you need to.)
  댓글 수: 1
Sam Hurrell
Sam Hurrell 2023년 4월 4일
편집: Sam Hurrell 2023년 4월 4일
That's really helpful thanks, but how do I edit it to remove the top half of the graph (-90, 180, 90) leaving a semi-circular plot?

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